Question Number 179105 by Shrinava last updated on 24/Oct/22
$$\mathrm{1}.\:\mathrm{Compare}:\:\:\:\pi^{\mathrm{2022}\boldsymbol{\mathrm{e}}} \:\:\:\mathrm{and}\:\:\:\mathrm{e}^{\mathrm{2022}\boldsymbol{\pi}} \\ $$$$\mathrm{2}.\:\mathrm{Compute}\:\mathrm{value}\:\mathrm{of}\:\:\:\mathrm{P}\:=\:\pi^{\boldsymbol{\pi}^{\boldsymbol{\pi}^{…^{\boldsymbol{\pi}} } } } \\ $$
Answered by MJS_new last updated on 25/Oct/22
![1. same as comparing Ο^e and e^Ο Ο^e <>^(?) e^Ο e ln Ο <>^(?) Ο ln e (e/(ln e))<>^(?) (Ο/(ln Ο)) ((d[(x/(ln x))])/dx)=(1/(ln x))β(1/(ln^2 x)) { ((<0; 0<x<e)),((=0; x=e)),((>0; x>e)) :} Ο>e β (e/(ln e))<(Ο/(ln Ο)) β e ln Ο <Ο ln e β Ο^e <e^Ο β β Ο^(2022e) <e^(2022Ο)](https://www.tinkutara.com/question/Q179113.png)
$$\mathrm{1}.\:\mathrm{same}\:\mathrm{as}\:\mathrm{comparing}\:\pi^{\mathrm{e}} \:\mathrm{and}\:\mathrm{e}^{\pi} \\ $$$$\pi^{\mathrm{e}} \overset{?} {<>}\mathrm{e}^{\pi} \\ $$$$\mathrm{e}\:\mathrm{ln}\:\pi\:\overset{?} {<>}\:\pi\:\mathrm{ln}\:\mathrm{e} \\ $$$$\frac{\mathrm{e}}{\mathrm{ln}\:\mathrm{e}}\overset{?} {<>}\frac{\pi}{\mathrm{ln}\:\pi} \\ $$$$\frac{{d}\left[\frac{{x}}{\mathrm{ln}\:{x}}\right]}{{dx}}=\frac{\mathrm{1}}{\mathrm{ln}\:{x}}β\frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \:{x}}\:\begin{cases}{<\mathrm{0};\:\mathrm{0}<{x}<\mathrm{e}}\\{=\mathrm{0};\:{x}=\mathrm{e}}\\{>\mathrm{0};\:{x}>\mathrm{e}}\end{cases} \\ $$$$\pi>\mathrm{e}\:\Rightarrow\:\frac{\mathrm{e}}{\mathrm{ln}\:\mathrm{e}}<\frac{\pi}{\mathrm{ln}\:\pi}\:\Rightarrow\:\mathrm{e}\:\mathrm{ln}\:\pi\:<\pi\:\mathrm{ln}\:\mathrm{e}\:\Rightarrow\:\pi^{\mathrm{e}} <\mathrm{e}^{\pi} \:\Rightarrow \\ $$$$\Rightarrow\:\pi^{\mathrm{2022e}} <\mathrm{e}^{\mathrm{2022}\pi} \\ $$
Commented by Acem last updated on 25/Oct/22
$$\pi>\mathrm{e}\:\mid_{{it}'{s}\:{ok}} \:\Rightarrow\:\frac{\mathrm{e}}{\mathrm{ln}\:\mathrm{e}}\overset{{how}?} {<}\frac{\pi}{\mathrm{ln}\:\pi}\: \\ $$
Commented by MJS_new last updated on 25/Oct/22
$$\frac{{x}}{\mathrm{ln}\:{x}}\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{for}\:{x}>\mathrm{e}\:\mathrm{because}\:\mathrm{its}\:\mathrm{derivate} \\ $$$$\mathrm{is}\:>\mathrm{0}\:\mathrm{as}\:\mathrm{I}\:\mathrm{showed} \\ $$
Answered by MJS_new last updated on 25/Oct/22
$$\mathrm{2}.\:{P}=+\infty \\ $$